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đề kiểu gì vậy ?

\(A=x^2-2x.\frac{2^2}{x}-5x+2\)

\(A=x^2-2.2^2-5x+2\)

\(A=x^2-8-5x+2\)

\(A=x^2-5x+2-8\)

\(A=x^2-5x-6\)

\(B=x^2-64.\frac{x}{2}-2x-80\)

\(B=x^2-32x-2x-80\)

\(B=\left(x-32-2\right)x-80\)

\(B=\left(x-30\right)x-80\)

\(B=x^2-30x-80\)

Chúc bạn học tốt

\(\frac{x^2-x-xy+y}{xy-x-y^2+y}\)

ĐKXĐ : \(\hept{\begin{cases}x\ne y\\y\ne1\end{cases}}\)

\(=\frac{\left(x^2-x\right)-\left(xy-y\right)}{\left(xy-x\right)-\left(y^2-y\right)}\)

\(=\frac{x\left(x-1\right)-y\left(x-1\right)}{x\left(y-1\right)-y\left(y-1\right)}\)

\(=\frac{\left(x-1\right)\left(x-y\right)}{\left(y-1\right)\left(x-y\right)}\)

\(=\frac{x-1}{y-1}\)

Bài làm 

\(\frac{x^2-x-xy+y}{xy-x-y^2+y}=\frac{x\left(x-1\right)-y\left(x-1\right)}{x\left(y-1\right)-y\left(y-1\right)}=\frac{\left(x-1\right)\left(x-y\right)}{\left(x-y\right)\left(y-1\right)}\)

\(=\frac{x-1}{y-1}\)

\(A=x^2-2x+y^2-4y+7=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+2\)

\(=\left(x-1\right)^2+\left(y-2\right)^2+2\)

Vì : \(\left(x-1\right)^2\ge0\forall x;\left(y-2\right)^2\ge0\forall y\)

=)) \(\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\)

Vậy GTNN A = 2 <=> x = 1 ; y = 2

\(A=x^2-2x+y^2-4y+7\)

\(=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+2\)

\(=\left(x-1\right)^2+\left(y-2\right)^2+2\)

Vì \(\left(x-1\right)^2\ge0\forall x\); \(\left(y-2\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\forall x,y\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-1=0\\y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}\)

Vậy \(minA=2\)\(\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}\)

Q=bc/a +ac/b+ab/c

Q=abc/a²+abc/b²+abc/c²

Q=abc x (1/a²+1/b²+1/c²)

Q=8 x 3/4

Q=6

y² + 2(x² + 1) = 2y(x + 1)

=> y² + 2x² + 2 = 2xy + 2y

=> y² + 2x² + 2 - 2xy - 2y = 0

=> y² - 2xy + x² - 2y + 2x + 1 + x² - 2x + 1 = 0

= (y - x)² - 2(y - x) + 1 + (x - 1)² = 0

=> (y - x - 1)² + (x - 1)² = 0

=> \(\hept{\begin{cases}y-x-1=0\\x-1=0\end{cases}}\Rightarrow\hept{\begin{cases}y=2\\x=1\end{cases}}\)

Vậy x = 1  ; y = 2 là giá trị cần tìm

Ta có P = a² + ab + b² - 3a - 3b + 1989

= \(\left(a^2+ab+\frac{1}{4}b^2\right)-3a+\frac{3}{2}b+\frac{9}{4}+\frac{3}{4}b^2-\frac{9}{2}b+\frac{27}{4}+1980\)

= \(\left(a^2+ab+\frac{1}{4}b^2\right)-3\left(a-\frac{1}{2}b\right)+\frac{9}{4}+\frac{3}{4}b^2-3.\frac{3}{2}.2\frac{1}{2}b+\frac{27}{4}+1980\)

= \(\left(a+\frac{1}{2}b\right)^2-3\left(a-\frac{1}{2}b\right)+\frac{9}{4}+3\left(\frac{1}{4}b^2-2.\frac{3}{2}.\frac{1}{2}b+\frac{9}{4}\right)+1980\)

= \(\left(a+\frac{1}{2}b-\frac{3}{2}\right)^2+3\left(\frac{1}{2}b-\frac{3}{2}\right)^2+1980\ge1980\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}a+\frac{1}{2}b-\frac{3}{2}=0\\\frac{1}{2}b-\frac{3}{2}=0\end{cases}}\Rightarrow\hept{\begin{cases}a+\frac{1}{2}b=\frac{3}{2}\\b=3\end{cases}}\Rightarrow\hept{\begin{cases}a=0\\b=3\end{cases}}\)

Vậy Min P = 1980 <=> a = 0 ; b = 3

Sửa đề: \(\frac{\left|x\right|-3}{x^2-9}\)

Ta có : \(\left|x\right|=\orbr{\begin{cases}-x\\x\end{cases}}\)

TH1 : \(\frac{-x-3}{x^2-9}=\frac{-\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{-1}{x-3}\)

TH2 : \(\frac{x-3}{x^2-9}=\frac{x-3}{\left(x-3\right)\left(x+3\right)}=\frac{1}{x+3}\)

Áp dụng bất đẳng thức Bunyakovsky dạng phân thức: \(\frac{x^2}{y-1}+\frac{y^2}{x-1}\ge\frac{\left(x+y\right)^2}{\left(x+y\right)-2}\)

Ta cần tìm giá trị nhỏ nhất của \(\frac{\left(x+y\right)^2}{\left(x+y\right)-2}\)

Đặt \(x+y=t\)thì ta có: \(\left(t-4\right)^2\ge0\forall t\Leftrightarrow t^2\ge8t-16\Leftrightarrow\frac{t^2}{t-2}\ge8\)

Vậy MinA = 8 khi và chỉ khi x = y = 2